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Proof step 14 from proof max-a (previous step)

n < m, n < n0, n0 ≤ m ⊦ m = max(max(m, n), n0), max(m, n) < n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
n < n0, n < m, n < max(m, n), n0 ≤ m, n0 ≤ max(m, n) ⊦ m = max(m, n)

Rule argument

Additional information

Used simplifier rules: Max-right: ⊦ ¬ m < n → max(m, n) = m;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)